From a building,two balls A and B are thrown | vedclass

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(C) Using the third equation of motion: $v^2 = u^2 + 2gh$,where $u$ is the initial speed,$g$ is the acceleration due to gravity,and $h$ is the height

Vedclass · Accepted Answer

$v_{A} = v_{B}$

Vedclass · Answer

(C) Using the third equation of motion: $v^2 = u^2 + 2gh$,where $u$ is the initial speed,$g$ is the acceleration due to gravity,and $h$ is the height of the building.For ball $A$ thrown upwards: The initial velocity is $u$ upwards. It will rise to a certain height and then fall back to the ground level. When it passes the point of projection on its way down,its speed will be $u$ downwards. Thus,it reaches the ground with a speed $v_A = \sqrt{u^2 + 2gh}$.For ball $B$ thrown downwards: The initial velocity is $u$ downwards. It reaches the ground with a speed $v_B = \sqrt{u^2 + 2gh}$.Since both $u$,$g$,and $h$ are the same for both balls,we have $v_A = v_B$.

From a building,two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ is thrown downwards with the same speed $u$ (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground,then:

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